answer:1. **Calculate the slope ( m )**: The slope ( m ) of a line through points ((x_1, y_1) = (0, 3)) and ((x_2, y_2) = (2, -1)) is calculated as: [ m = frac{y_2 - y_1}{x_2 - x_1} = frac{-1 - 3}{2 - 0} = frac{-4}{2} = -2 ] 2. **Interpret the y-intercept ( b )**: Since the line passes through ((0, 3)), the y-intercept ( b ) is ( 3 ). 3. **Calculate the product ( mb )**: [ mb = m cdot b = (-2) cdot 3 = -6 ] 4. **Determine the correct range for ( mb )**: Since ( mb = -6 ), it falls in the range ( mb < -1 ). Conclusion: The correct statement about ( mb ) is that it is less than -1. Thus, the final answer is mb < -1. The final answer is boxed{A}

answer:1. **Initial Setup**: Let the original price be P and original demand be D. The new price P' after a 20% increase is 1.20P. The quality improvement is assumed to potentially increase demand by 10%, but actual demand change needs to be calculated under the new conditions. 2. **Calculating Demand Change**: If quality alone could increase demand by 10%, the potential new demand D_q is 1.10D. However, the actual demand D' must counteract both the increase in price and the change due to quality. To maintain revenue, the condition is: [ P cdot D = P' cdot D' Rightarrow P cdot D = 1.20P cdot D' ] From this, we get: [ D' = frac{D}{1.20} ] 3. **Integrating Quality Effect**: We need to find D' such that it considers the quality improvement. Assume D' proportionally adjusts between D_q and frac{D}{1.20}: [ D' = frac{1.10D}{1.20} = frac{11}{12}D ] Therefore, the percentage change in demand is: [ frac{D' - D}{D} cdot 100% = left(frac{11}{12} - 1right)D cdot 100% = -frac{1}{12} cdot 100% = -8.33% ] So, the demand needs to decrease by approximately 8.33% to keep the revenue at least constant. Conclusion: The required change in demand, considering both the price increase and quality improvement, is a decrease of approximately -8.33%. The final answer is boxed{text{C}}

answer:1. **Utilize the South Pole Theorem**: Remember that the bisector of widehat{ABC} intersects Gamma at the midpoint M of the arc AC. This M is the center of a circle passing through B, C, and the center of the inscribed circle of triangle ABC. The proof of this theorem involves basic angle chasing, and we'll use it to construct our solution. 2. **Identify Key Points**: Let K and L be the midpoints of arcs AB and AC respectively. By the South Pole Theorem, points P, I, and K are collinear, as are points P, J, and L. Moreover, KA = KB = KI and LA = LC = LJ because K and L are on the perpendicular bisectors of AB, AC respectively. 3. **Similarity Construction**: Now, consider the second intersection point we need to find, denoted Q. By our construction, Q is the center of the direct similarity that maps K to L and I to J. We need to demonstrate that the similarity which transforms K to L and I to J remains invariant as P varies along the arc BC. 4. **Angle and Ratio Analysis**: The angle of similarity is given by the angle between the lines IK and JL, denoted by angle(KA, AL), which is fixed because K, A, and L are fixed points on the circle. Furthermore, considering the lengths, the ratio of the lengths frac{JL}{IK} = frac{AL}{AK} is also fixed as A, K, and L are fixed points determining these segments. 5. **Conclusion on Similarity**: Given that both the angle and ratio of the similarity are fixed, there’s only one such similarity that can map K to L and I to J. It follows that the similarity’s center, which is the point Q, is fixed for any position of P on the arc BC. Thus, Q is static. In conclusion, the fixed second intersection of the circumscribed circle of triangle PIJ and Gamma confirms that the position of Q remains unchanged. blacksquare

answer:**Step 1: Analyze the problem** This problem involves the relationship between trigonometric functions, double-angle formulas, and the graph and properties of the sine function. According to the question, using trigonometric identities, we have f(x)=dfrac{1}{2}sin 2x+1. Then, by utilizing the graph and properties of the sine function, we can obtain the result. **Step 2: Determine the expression for f(x)** f(x)=sin xcos x+(1+tan^{2}x)cos^{2}x=dfrac{1}{2}sin 2x+left(1+dfrac{sin^{2}x}{cos^{2}x}right)cos^{2}x=dfrac{1}{2}sin 2x+cos^{2}x+sin^{2}x=dfrac{1}{2}sin 2x+1. **Step 3: Calculate the minimum positive period** Since f(x)=dfrac{1}{2}sin 2x+1, the period of sin 2x is dfrac{2pi}{2}=pi. As the additional constant 1 does not affect the period, the minimum positive period of the function is T=pi. **Step 4: Calculate the maximum value** The maximum value of sin 2x is 1. Thus, the maximum value of dfrac{1}{2}sin 2x is dfrac{1}{2}. Consequently, the maximum value of the function is dfrac{1}{2}+1=boxed{dfrac{3}{2}}.